Question: Divide the following complex numbers. $\dfrac{-32+8i}{5+3i}$
Solution: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate, which is ${5-3i}$. $ \dfrac{-32+8i}{5+3i} = \dfrac{-32+8i}{5+3i} \cdot \dfrac{{5-3i}}{{5-3i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$. $ = \dfrac{(-32+8i) \cdot (5-3i)} {5^2 - (3i)^2} $ Evaluate the squares in the denominator and subtract them. $ = \dfrac{(-32+8i) \cdot (5-3i)} {(5)^2 - (3i)^2} $ $ = \dfrac{(-32+8i) \cdot (5-3i)} {25 + 9} $ $ = \dfrac{(-32+8i) \cdot (5-3i)} {34} $ The denominator now doesn't contain any imaginary unit multiples, so it is a real number. Note that when a complex number, $a + bi$ is multiplied by its conjugate, the product is always $a^2 + b^2$. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-32+8i}) \cdot ({5-3i})} {34} $ $ = \dfrac{{-32} \cdot {5} + {8} \cdot {5 i} + {-32} \cdot {-3 i} + {8} \cdot {-3 i^2}} {34} $ $ = \dfrac{-160 + 40i + 96i - 24 i^2} {34} $ Finally, simplify the fraction. $ \dfrac{-160 + 40i + 96i + 24} {34} = \dfrac{-136 + 136i} {34} = -4+4i $